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Sequential Multiple Hypothesis Testing with Type I Error Control
A
Proofs
Proof of Lemma 4.1. Let Tb be the smallest t at which
the positive martingale Λt ≥ b, with Tb = ∞ if the
threshold is never crossed. Then Tb is a stopping time.
For some number of visitors t, we have
(
b Tb ≤ t,
Λmin{Tb ,t} ≥
0 Tb > t.
Therefore, we may write
1 = E[Λmin{Tb ,t} ] ≥ bP (Tb ≤ t),
where the equality comes from the optional stopping
theorem. Taking t → ∞ bounds the probability of ever
crossing the threshold at P (Tb ≤ ∞) ≤ 1/b, which
means P (∃ t such that Λt ≥ b) ≤ 1/b. This completes
the proof of this lemma.
Proof of Corollary 4.2. Based on fundamental arguments in probability theory, we have the following
chain of inequalities:
1
0
P
≤ δ = P max
Λn ≥ 1/δ
n0 ≤t
maxn0 ≤t Λn0
To
R start with, since the null hypothesis is simple,
P (θ0 |M0 )P (Dn |θ0 , M0 )dθ0 = P (Dn |M0 , θ0 ). Hence
we may write the Bayes factor as
R
Pr(θ1 |M1 ) Pr(Dn |θ1 , M1 )dθ1
Λn = R
Pr(θ0 |M0 ) Pr(Dn |θ0 , M0 )dθ0
R
Qn
P (θ1 |M1 ) t=1 P (xt |θ1 , M1 )dθ1
Qn
=
t=1 P (xt |θ0 , M0 )
The expectation under the null hypothesis H0 from Λn
conditioned on whatever observed up to time n − 1,
i.e., Fn−1 , we have
E0 [Λn |Fn−1 ]
#
"Z
Qn−1
P (θ1 |M1 )P (xn |θ1 , M1 ) t=1 P (xt |θ1 , M1 )
dθ1
= E0
Qn−1
P (xn |M0 ) t=1 P (xt |M0 )
Qn−1
Z
P (xn |θ1 , M1 )
t=1 P (xt |θ1 , M1 )
= P (θ1 |M1 ) E0
dθ1
Q
n−1
P (xn |M0 )
t=1 P (xt |M0 )
{z
}
|
=1
Z
=
R
=
Qn−1
t=1 P (xt |θ1 , M1 )
dθ1
P (θ1 |M1 ) Q
n−1
t=1 P (xt |M0 )
P (θ1 |M1 )P (Dn−1 |θ1 , M1 )dθ1
= Λn−1
P (Dn−1 |M0 )
≤ P (∃ t such that Λt ≥ 1/δ)
≤δ
where the last step follows from Lemma 4.1. This
completes the proof of this corollary.
Proof of Corollary 4.3. We first show that the likelihood ratio is a positive martingale under the simple
null hypothesis H0 : θ0 . If we take the expectation
under the null hypothesis H0 from the likelihood ratio
Λt , i.e.,
Qn
Pr(Dt |θ1∗ ) i.i.d. t=1 P (xt |θ1∗ )
Qn
Λt =
=
∗
Pr(Dt |θ0∗ )
t=1 P (xt |θ0 )
conditioned on whatever observed up to time t − 1,
i.e., Ft−1 , we have
t−1
P (xt |θ1∗ ) Y P (xt |θ1∗ )
E0 [Λt |Ft−1 ] = E0
P (xt |θ0∗ ) t=1 P (xt |θ0∗ )
Z
P (xt |θ1∗ )
= Λt−1
P (xt |θ0∗ )dxt
P (xt |θ0∗ )
{z
}
|
Proof of Lemma 5.5. Recall that m0 is the number of
true null hypotheses which we assume have indices
0
1, . . . , m0 ; thus, p1T , . . . , pm
T are all sequential p-values.
This implies that there must exist an increasing function g k with g k (x) ≤ x such that Pr(g k (pkT ) ≤ δ) = δ;
thus, g k (pkT ) ∼ Uniform[0, 1]. If pkT is discrete, then we
may need to allow g to be random, e.g. g k (pkT ) + ξ,
where ξ is chosen to interpolate between subsequent
values of pkT .
Define V = P g 1 (p1T ), . . . , g m (pm
t ) ∩ {1, . . . , m0 } as
the number of true hypotheses rejected by P on the
modified p-values. We will argue that VT ≤ V almost
surely.
First, suppose that P is a step-up procedure and V = i;
this implies that g (k) (p(k) ) ≤ αi for all k ≤ i and
αi < g (k) (p(k) ) for all k > i. Since g(p) ≤ p, we must
have
αi < g (k) (p(k) ) ≤ p(k) ∀k < i
where E0 is the expectation under the null hypothesis
H0 .
and hence Vt ≤ V a.s.. This implies that E[f (VT , s)] ≤
E[f (V, s)], and using the the guarantee of P on f (V, s)
implies E[f (VT , s)] ≤ E[f (V, s)] ≤ q for all s, which
yields the theorem statement by linearity of expectation.
We then show that the Bayes factor is also a
test martingale if the null hypothesis is simple.
If P only have an independent guarantee, note
0
that if p1T , . . . , pm
are independent, then so are
T
=1
= Λt−1
Alan Malek, Sumeet Katariya, Yinlam Chow, and Mohammad Ghavamzadeh
0
g 1 (p1T ), . . . , g m0 (pm
T ) and we can apply the same reasoning as above to imply the (f, q) guarantee in the
independent case.
Proof of Lemma 5.6. Let M = R(p1T1 , . . . , pm
Tm ),
M0 = R(p1N , . . . , pm
)
be
the
tests
rejected
by
S(P)
N
and S 0 (P) and R = |M|, R0 = |M0 | be their cardinally,
respectively.
Consider the case when P is a step-up procedure. We
show that R = R0 . The monotonicity of the sequential
p–values implies that pkTk ≤ pkN for all k, and thus
R ≤ R0 ; if pkTk is rejected, then pkN must be as well. We
also argue that R ≥ R0 ; suppose, instead, that R < R0 ,
(R)
(R0 )
which implies that pT(R) ≤ αR < pN ≤ αR0 , which is
a contradiction since Tk = N for all k that were not
already stopped. Hence, R = R0 .
If the two procedures both reject R tests, then αR <
(V +1)
(m)
(R+1)
(m)
pN
, . . . , pN and αR < pT(R+1) , . . . , pT(m) correspond to the tests that have not been rejected. But we
have T(R+1) , . . . , T(m) = N , so M = M0 .
Now consider the case when P is a step-down proce(1)
dure. If R tests are rejected by S(P), then pT(1) ≤
(R)
(k)
(k)
α1 , . . . , pT(R) ≤ αR . Since pN ≤ pT(k) , we must have
M ⊆ M0 . Now, suppose there exists an index k such
that test k is rejected by S 0 (P) but not S(P). This
implies that pkN ≤ αR0 but pkTk > αR0 . However, by
construction of Tk , we have that Tk < N only if test
k is rejected, which implies that pkTk > αR0 cannot
happen. Thus, we have M = M0 for the step-down
case as well.